Lesson video

Welcome, and well done for making the decision to learn using this video today.

My name is Ms. Davies, and I'm gonna be helping you as you work your way through this lesson.

There's lots of opportunities for you to really think today and test your understanding.

Make sure you are pausing the video to give yourself that chance to think.

And then I'll help you using any hints and tips as we go through.

If you've got everything you need, then let's get started.

This lesson is called a single solution set.

By the end of the lesson, you'll be able to compare two solution sets from two different inequalities and define a solution set that satisfies both.

If you need a reminder as to what an equality is, pause video and read through that now.

We're gonna start then by looking at solution sets for multiple inequalities.

We can sometimes write the values that satisfy two inequalities as a single solution set.

What values satisfy x greater than 3 and x less than or equal to 4 simultaneously? All right, well, any values in that range.

We call this the set of values which satisfy both inequalities.

All right, what values satisfy these simultaneously? You'll notice that we've changed the inequality sign for our second inequality to x is greater than or equal to 4 this time.

If I put those on a number line, x has to be greater than 3 and greater than equal to 4.

So any values in the set where x is greater than or equal to 4 satisfied both inequalities.

We can do the same with inequalities with two constraints.

So what values satisfy both of these inequalities simultaneously? What do you think? Well, a number line might help us.

So if I draw the first equality on the number line and the second, we want values that satisfy both.

So we're looking at the overlap with our two inequalities.

So it's any value greater than or equal to -1 and less than or equal to 2.

What is the problem with finding the set of values which satisfy both of these inequalities? Have a read.

What do you think? All right, if we draw them on a number line, we can see that there's no overlap.

If we look at 2, the first inequality says we have to be less than 2, the second inequality says we have to be greater than 2, which is why neither are shaded.

So 2 is not included in either.

It's impossible for a number to be in both sets.

So there is no single solution set for both inequalities.

Now, it's also important to know if we're looking for values which have to satisfy both inequalities simultaneously or satisfy either inequality.

So children under 5 or adults 60 and above can ride the bus for free.

How could we represent this as a solution set? What do you think? All right, age cannot be negative.

So we could show this as x is greater than or equal to 0 but less than 5 and greater than or equal to 60.

There's it on a number line.

Clearly, it's not possible to satisfy both of these inequalities simultaneously.

There's no overlap.

So we could write it with an or statement.

So in context, x can be greater than or equal to 0 and less than 5 or x is greater than or equal to 60.

We're not solving those simultaneously.

There's no values in both, but there could be a value in one or the other.

All right, how can we write all values which satisfy both of these inequalities? Give this one a go.

Always use a number line to help you.

We want values greater than or equal to 0 and less than 3.

So well done if you picked C.

So we can use this skill to find the solution set which satisfies multiple inequalities, and this is the bit that might be new to you.

We're gonna find the solutions which satisfy both of these.

So we can start by solving one.

So we get 2x is greater than 2, x is greater than 1.

Any value greater than 1 satisfies that first inequality.

We can put it on a number line to help us.

And then we can solve the other one.

So we get x is less than or equal to 5.

In order to satisfy both, the solution will have a value greater than 1 but less than or equal to 5.

So that's our solution set that satisfies both inequalities.

Let's try a different example.

Let's find the solutions which satisfy both of these.

So again, if we apply our rules of algebraic manipulation to our first equality, and then the same with our second inequalities.

So we need to add 1, and then divide by 5 over 3, which is the same as multiplying by the reciprocal.

So we could do 5 multiplied by 3 over 5, which is x is greater than or equal to 3.

All right, if we draw 'em both on the number line, we can see that our constraints are both acting in the same direction.

If we need values that both, we need to look at the larger value.

So x has gotta be greater than 4 to satisfy both of those.

Sometimes there are no values which satisfy two different inequalities.

What happens when we solve these? You might wanna pause the video and have a go at solving these first.

Looking at the first one, you could expand the brackets.

A quicker way will be to divide by 3, then add 3, then half.

You should get x is greater than or equal to 4.

Let's look at the other one.

Subtracting 4, divided by 8, x is less than 2.

Now, of course, there are no values that satisfy both here, let's draw them on our number line.

So there's no overlap, no solutions to both inequalities simultaneously.

And it would not make sense to write this as an or statement 'cause we're looking for a value which satisfies both, and that are no values that satisfy both, All right, which of these are solutions to both of these inequalities? Okay, well you could substitute them into check, but a quicker way would be to solve the inequalities.

So you've got x is less than 6, then divide by 3, and add 2, x is greater than or equal to 3.

So we want values greater than or equal to 3, but less than 6.

Time to put that into practise.

For each question, I'd like you to write the solution set for both inequalities simultaneously.

And so you've got two to do from number lines and then two to do from inequalities.

Give those a go.

Question two, same idea.

I'd like you to find the range of values that satisfy both inequalities, but you're gonna have to do a little bit more to solve these this time.

Be really careful with your steps of algebraic manipulation.

Make sure you present your working really neatly so that you can go back through and check.

When you're happy with those, come back for the next bit.

Question three, Lucas has tried to find the range of solutions which satisfied both of these inequalities.

He says, "Solutions which satisfy both are when x is less than or equal to -3." What mistake has he made? And then, can you substitute a value from Lucas's solution to show him it is not valid? When you happy with your answers, come back and we'll look at the last bit.

And finally, for each pair of inequalities, state whether it is possible to find the values which satisfy both.

If it's possible, write the set of solutions which satisfy both.

If not, you can just write, "No solutions." Give those ones a go.

Let's have a look.

So for A, x is greater than or equal to -4 but less than 0.

For B, x is greater than or equal to -2 and less than or equal to 1.

For C, x is greater than or equal to 2 and less than or equal to 4.

D, x is greater than or equal to -2 and less than 5.

Make sure you've got less than 5 'cause that second inequality says we have to be less than 5, so we can't include 5.

For two, I'd like to pause the video and just check your lines of working, and then check your final solution.

When you're happy with those, we'll look at the next set.

Question three, so Lucas's mistake was at the very end.

He's done most of it correctly.

His values have to be less than or equal to -3 and less than -5.

So in order to satisfy both of those has to be less than -5.

So sort of values we need to pick to show him it's incorrect are values -3 or less, which are greater to the -5.

So if you're looking at integer values, that's -3 and -4.

Of course, you could use a non-integer value in that range as well.

I picked -4.

So I've got 15 is less than 5.

Subtract 2 dots of -4.

Because 15 is less than 5 plus 8, which gives you 15 is less than 13.

Well, that's not true.

So a solution must be incorrect.

And we've said why that occurred in the first part.

We're actually looking for values less than -5.

And then finally, pause video and check your answers to these.

There are no solutions to A, solutions to B have to be greater than 0 but less than 1, solutions to C, any value less than 20.

Please take your time to read through those steps are working and then we'll move on and look at the second part of the lesson.

So now, I'm gonna look at solving and combining more complex inequalities so we can solve more complicated inequalities and combine the solutions.

We'll do this one together.

What are the solutions to these inequalities simultaneously? So let's look at that first inequality.

Now, we do not need to separate this into two inequalities, we can just apply the same operations to all expressions, and that's gonna be more efficient.

So here we can divide all expressions by 5.

So 3 is less than or equal to 2x minus 3, which is than 9.

Then we can add 3, and divide by 2.

So x is greater than or equal to 3 but less than 6.

I'm gonna put that solution so far on a number line.

And then don't forget, we've got the second inequality we need to satisfy as well.

So if we add 3, divide by 4/5, which is the same as multiplying by 5 over 4.

And that gives us x is less than or equal to 5.

Putting that on a number line, we can clearly see our solutions are when x is greater than or equal to 3 and less than or equal to 5.

Izzy says, "What would happen if I changed the second inequality to 4x over 5 minus 3 is greater than or equal to 1?" So changing the inequality sign in that second inequality.

Why don't you pause the video and think about how that will affect our solution.

Well, the number line is gonna be really helpful here because our solution is now gonna be when x is greater than or equal to 5.

If we want a solution in both our inequalities, then its values greater than or equal to 5 but less than 6.

With some inequalities, it may be necessary to separate them into multiple inequalities to solve simultaneously.

So we're gonna look at finding solutions to these inequalities.

So that first equality we can separate into two different inequalities.

We've got 5 minus 2x is greater than or equal to x plus 2, and 5 minus 2x is less than 8 minus x.

And see how I'm presenting this really clearly so I don't get in a model as to which inequality I'm looking at.

And then I've also got my other inequality, so I'm gonna write that out as well.

So let's look at this first inequality, adding 2x, subtracting 2, divided by 3.

X is less than or equal to 1.

Let's look at the other inequality.

Adding 2x, subtracting 8, and then we've got x is greater than -3.

And the last one, subtracting 4 and divided by 3, x is less than 0.

And then all we need to do is find solutions which satisfy all of those constraints.

What are they gonna be? I've put them all on a number line to help me.

And you can see, they're gonna be values greater than -3 but less than 0.

That other constraint is not needed now.

And that's how we'd write it as an inequality.

Izzy says, "I want to find the solutions to these inequalities simultaneously.

But they look really tricky!" What could Izzy do first? Jacob says, "It may look tricky, but we can just solve one inequality at a time." So it doesn't matter which one you chose to do first.

I've gone with 3x plus 5 is greater than or equal to 8.

I've just looked at that lower constraint of the first inequality.

Solving that gives me x is greater than or equal to 1.

But I also know that 3x plus 5 has to be less than 5x minus 11.

And again, we can solve that to get x is greater than 8.

I'm not worried about my solution set at the moment.

I'm gonna solve my last equality first.

And this one, I don't need to separate into two inequalities.

I can just multiply all expressions by 3, add 9, and divide by 2.

And you can see now how important it is to keep our working out really clearly down the page.

Izzy says, "The solution is the set of numbers where x is greater than 3 but less than 8." Do you agree? Jacob says, "I'm not sure, I think I need a number line to help," and that's absolutely fine.

Now, if we're drawing our own number line, we don't necessarily want to put all the lines on and put the numbers on the lines because we don't know exactly what values we're going to need.

So we're not gonna do this to scale, we're just gonna do this roughly.

So x has got to be greater than or equal to 1.

Now, 1 is the smallest number that we've got in our solution.

So I'll put that down to the left.

We've got x is greater than or equal to 1.

Then we've got x is greater than 8.

So if we go a little bit further along and put an 8, again, doesn't matter if we're to scale.

And then we have x is greater than the 3 but less than 18.

So three is roughly there, 18 is gonna be further up by number line.

And now, I want values that satisfy all of these.

All right, Izzy was incorrect.

I think she's probably misread her second inequality.

X has to be greater than 8, not less than 8.

So we need values where x is greater than 8 but less than 18.

There's our solution set there.

All right, I'm gonna show you one on the left-hand side and then you are gonna have a go on the right-hand side.

You've seen lots of examples now, so you should be really good at this.

So the first equality, I'm gonna split into two different inequalities.

So in the left-hand one, gives me x is greater than or equal to -1, and our other constraint, x is less than 2.

I'd like you to get to the same stage with your first inequality.

So we need to split that into two inequalities.

So we've got 3x plus 1 is greater than 2x.

Subtracting 2x gives us x plus 1 is greater than 0.

So x is greater than -1.

Looking at the other constraint, 2x plus 1 is less than 15.

We've subtracted x from both sides.

Then, we need to subtract 1 and divide by 2.

And we get x is less than 7.

All right, if we look back at my example, I'm now gonna look at my second inequality.

So 5x plus 20 is greater than or equal to 25.

5x is greater than or equal to 5.

So x is greater than or equal to 1.

How about you to solve your second inequality as well? So if we add 5, we get 4x is less than 20, x is less than 5.

And now all we need to do is combine our solutions.

So I've sketched myself in number line to help me.

<v ->1 is my smallest value, let's put it down there,</v> and I've got x is greater than or equal to -1.

I need to make sure my arrows go in the right way.

And then x is less than 2.

And then x is also greater than or equal to 1.

There we go.

And now I can clearly see that to satisfy all three, x has gotta be greater than or equal to 1 but less than 2.

Have a go on your one.

What is your final solution set? You've got x is greater than -1 and less than 7 but also less than 5.

Your final solution set x is greater than -1 but less than 5.

Fantastic! Time to put that all into practise.

There's nothing new here, it's just the skill of solving an inequality and then piecing it together at the end.

However, there's lots of room to make mistakes with your algebra manipulation, so make sure that you are laying things out really clearly.

I can't stress enough how important that is when you're looking at these trickier algebra manipulations.

I'd also like you to represent your solution on a number line.

It doesn't have to be to scale, but you need to mark on any values which you are using.

Give those two a go.

Question three, I'd like you to do exactly the same.

I'd like you to write the set of values which satisfy both inequalities simultaneously.

Then, which of those values are in the solution set? And then I've changed the inequality symbol in that second inequality.

What would the set of solutions be for those inequalities simultaneously? There's quite a lot to do there.

Give that your best shot and then we'll look at the next bit together.

For question four, I'd like you to write an inequality to show the set values which satisfy these simultaneously.

Give that one a go.

Question five, a little bit of a puzzle.

I'd like you to pick two inequalities from the options below which satisfy the given constraints, and then explain your answer.

So I want you to pick two outta those options, A, B, C, or D, so that the inequalities have no solutions when solved simultaneously.

Then pick two so that the inequalities have one solution when solved simultaneously.

And then pick two so the inequalities have solutions only when x is greater than -1 but less than 1.

You can use each equality more than once.

That's absolutely fine.

When you've played around with that puzzle, we'll look at the answers together.

Let's have a look.

So we can solve the first inequality by subtracting 10 and dividing by 3.

We don't need to separate it.

We get x is greater than or equal to -2 and less than or equal to 4.

And the second inequality, x is less than 1.

That means x has to be greater than or equal to -2 but less than 1.

And I've sketched that on my number line.

You'll see that I've marked on the -2 and the 1.

I've got a closed circle at -2 and an open circle at 1 and a line connecting them.

For 2a, again, we don't need to separate them into multiple inequalities.

We can just add 1 and divide by 2.

So we get x is greater than or equal to -2 and less than or equal to 7.

For the second inequality, you might want to separate those into two inequalities 'cause I've got a -x to deal with.

I've decided not to.

So I've subtracted 10 from each expression, but then I've got -x is greater than -3 but less than 5.

So I need to divide or multiply by -1.

Now, that means my inequality signs are going to change.

I've just gotta be careful with that.

So x is less than 3 but greater than -5.

And then I've written that the other way round because it makes more sense.

So x is greater than -5 but less than 3.

If you are not confident with that, dividing by -1 and changing the inequality signs, then split it into two separate inequalities, and start by adding x to both sides.

You'll get the same final answer.

So we've got x is greater than or equal to -2 but less than 3.

And I've drawn that on a number line.

Just check you've got the same as me and we'll look at the next bit.

The first inequality is not too bad to solve.

I can just subtract 2x from all three expressions, don't even need to separate it.

And then I can solve by subtracting 10 and divide by 3.

So the second one, you've got options with how you'd like to solve this.

I decided to put them over a common denominator.

So I've multiplied the numerator and the denominator of the first fraction by 2 and the numerator and the denominator at the second fraction by 3.

So they have a common denominator of 6.

That means that 6x plus 2 has to be less than 3x minus 6.

So then I can subtract 3x from both sides, subtract 2, and a divide by 3.

Solutions to both are when x is greater than -16 over 3 but less than -8 over 3.

So the values which are in both -2.

7 -4, and -15 over 3.

Little bit tricky there because you had a bit of fraction work to try and work out roughly what -16 over 3 and -8 over 3 was as a decimal.

So well done if you got those three.

If I change the inequality sign round, x would then have to be larger than -8 over 3.

So the lower constraint will be -8 over 3, the upper constraint is -2.

So x is greater than -8 over 3 and less than or equal to -2.

Don't forget that upper constraint from the first inequality.

All right, that was tough.

Well, I dunno if you got that final answer.

There was lots to do there.

For question four, pause video and check my stages of working.

There was quite a lot to do.

You should get a solution when x is greater than or equal to 8 but less than 15.

Then I wonder if you found these.

The easiest way to do this was to find the solution sets first and then look for the ones that satisfied these restrictions.

So the solution for the first inequality is when x is greater than -2 but less than 1.

Second inequality, x is greater than -1.

Third one, x is less than or equal to 2.

And then x is greater than or equal to 2.

So in order to have no solutions when solved simultaneously, you need A and D.

And that's because for A, we have to be less than 1, but for D, we need to be greater than or equal to 2.

So it's impossible to satisfy both.

For B, you should have C and D.

There's one solution to both simultaneously, and that's exactly when x is 2.

And finally, you should get A and B.

Solutions to both is when x is greater than -1 but less than 1.

Fantastic! You are now experts at solving inequalities and representing as a single solution set.

We've seen all sorts of examples, such as when our solutions have two constraints or one constraint, we've looked at when we might not have any solutions at all.

We've really played around with these inequalities.

Don't forget that a number line can be a really useful tool.

There's absolutely nothing with using those things to support you to get the correct answer.

Thank you for joining us today.

I hope you feel like you've really challenged yourself and improved on your inequality skills.

I really hope you choose to come back and learn with us again.